I want to show that there is a $1$-dimensional $C^\infty$-atlas for the topological boundary $M$ of $[0,1]^2$.$^1$
I tried to start with the "lower half" of $M$: Let $$\phi_1(x):=\begin{cases}(0,-x)^T&\text{, if }x\in(-1,0]\\(x,0)^T&\text{, otherwise}\end{cases}$$ and $$\phi_2(x):=\begin{cases}(1,-x)^T&\text{, if }x\in(1-x,0]\\(x,0)^T&\text{, otherwise}\end{cases}$$ for $x\in U_1:=U_2:=(-1,1)$. Note that $$N:=\phi_1(U_1)\cap\phi_2(U_2)=(0,1)\times\{0\}\tag1$$ and $$V_i:=\phi_i^{-1}(N)=(0,1).\tag2$$ Moreover, $$\phi_2^{-1}=(1-y_1,0)^T\;\;\;\text{for all }y\in N.$$
Question 1: Now my first problem is: Are the $(U_i,\phi_i)$ actually charts of $M$? I'd say that they lack differentiability at $0$. It's a subtlety, since I guess we can simply restrict both to $V_i$, right?
Question 2: How can we conclude that $M$ is a $C^\infty$-submanifold of $\mathbb R^2$?
$^1$ $(U,\phi)$ is called $1$-dimensional chart of $M$ if $U\subseteq\mathbb R$ is open, $\phi:U\to\mathbb R^2$ is an immersion and a topological embedding of $U$ into $M$ and $\phi(U)$ is $M$-open. If $\alpha\in\mathbb N_0\cup\{\infty\}$ and $(U_i,\phi_i)$ is a $1$-dimensional chart of $M$, then $(U_1,\phi_1)$ and $(U_2,\phi_2)$ are called $C^\alpha$-compatible if $N:=\phi_1(U_1)\cap\phi_2(U_2)=\emptyset$ or $\psi:=\phi_2^{-1}\circ\left.\phi_1\right|_{V_1}:V_1\to V_2$, where $V_i:=\phi_i^{-1}(N)$, is a $C^\alpha$-diffeomorphism. A $1$-dimensional $C^\alpha$ atlas $\mathcal A$ for $M$ is a collection of $C^\alpha$-compatible charts such that $M\subseteq\bigcup_{(U,\:\phi)\in\mathcal A}\phi(U)$.